Amy has a master's degree in secondary education and has taught math at a public charter high school.

Factoring is a quick and easy way to find the solutions to a quadratic trinomial. Watch this video lesson to learn how you can use this method to solve your quadratics.

A Quadratic Trinomial

First, what is a quadratic trinomial? In short, it is a quadratic expression with all three terms. What does this mean? If my quadratic expression is of the form ax^2 + bx + c, where a, b, and c are numbers, then my quadratic trinomial will make sure that neither a, b, nor c will be 0. All three of these letters will be a number other than 0.

Notice the little 2 next to the first x. That tells us that this expression is a quadratic because a quadratic means that your largest exponent is a 2. The trinomial part comes from the three parts or terms that make up our quadratic. Since none of our letters are 0, all of our terms are there and we have three parts separated by either a minus or a plus.

An example of a quadratic trinomial is 2x^2 + 6x + 4. Do you see how all three terms are present? All my letters are being represented by numbers. My a is a 2, my b is a 6, and my c is a 4.

Solving by Factoring

When we set our quadratic trinomial equal to 0, that's when we want to solve it. The method I'm going to show you in this lesson is how to solve by factoring. What this means is that we are finding what multiplies together to get to our quadratic.

After factoring, we will end up with two sets of parentheses, with each set being one of our factors. A quadratic will only have two sets. You can remember this by looking at the little 2, since that tells you how many sets or factors we will have.

Keep in mind that factoring is just one method of solving a quadratic. Other methods might be better for different quadratics.

Setting Up the Problem

To begin, we need to set up our problem so we can easily find our factors. We're going to solve 2x^2 + 6x + 4 = 0 by factoring. What I do is I first write out my problem and I have it equal two empty sets of parentheses like this: 2x^2 + 6x + 4 = ( . . . )( . . . ). I'm a visual person, so by writing this part out, I can clearly see that I need to fill both these parentheses with information.

My next step is to find the factors of my first term and my last term. I underline my first term and my last term.

First and last term.

Now I can write my factors underneath. For 2x^2, my factors are x and 2x. I write these under my 2x^2. For the 4, I can either have 1 and 4 or 2 and 2. Because the sign of the 4 is positive and two negatives multiplied together give a positive, I can also have -1 and -4 or -2 and -2 as possible factors. So I write these underneath the line for 4. I separate each group of factors with a line.

Factors of first and last term.

I am now done setting up my problem. The next step is to find my factors to help me solve.

Finding the Factor

My job is to combine a pair of factors from the first term with a pair of factors from the last term so that I get my middle term. Let me show you what I mean here. First, I look at the possible factors from the first term. I only have one pair to look at, the x and 2x. So I know I have to use these two. This first pair from the first term actually tells me what values go at the beginning of my parentheses. I can go ahead and fill that part in, like this: 2x^2 + 6x + 4 = (x . . )(2x . . ).

I now have to figure out what numbers I need to multiply with my x and 2x so that when I add them up I get 6x, the middle part. For these numbers, I look at the possible factors of my last term. I notice that the 6 is a positive, so that means I will have to add positive numbers together. So, I can scratch the -1 and -4 and the -2 and -2, since these will give me a negative number when added together after multiplying.

So that leaves me with either the 1 and 4 or the 2 and 2. I start playing with my first pair, the 1 and 4. I can combine them with the x and 2x in two ways. I can multiply the x with the 4 and the 2x with the 1 or I can multiply the x with the 1 and the 2x with the 4. Which way will give me a 6x when added together?

The first way works because x times 4 is 4x and 2x times 1 is 2x. Adding 4x and 2x gives 6x, my middle term. The other way doesn't work, since 2x times 4 is already 8x. Adding x times 1 or x to it gives us 9x, which isn't 6x. Since I've found my combination, I can stop looking. If I still haven't found my combination though, I would move on to the next possible pair of factors and continue until I've found it.

So now that I've found my combination, I need to finish filling in my parentheses. I know that my x has to multiply with the 4 and my 2x with the 1. Where do I put the numbers so that this will happen? A good way to remember is to tell yourself that parentheses multiply together, so your combinations must be in different parentheses.

My x is in the first pair of parentheses, so my 4 goes in the second. My 2x is in the second, so my 1 goes in the first. So now I have 2x^2 + 6x + 4 = (x + 1)(2x + 4). Since my 1 and 4 are positive, I am using the pluses. If my problem had a negative factor, then my sign for that factor would be a minus. I stop here if my problem only wanted me to factor. If it says to solve, then I need to continue.

To solve, I set each of my parentheses equal to 0. So, I have x + 1 = 0 and 2x + 4 = 0. Now I solve each for x and I will get my answers. And yes, I will have two answers or solutions. To solve x + 1 = 0, I need to subtract 1 from both sides and I get x = -1. To solve 2x + 4 = 0, I first need to subtract 4 from both sides to get 2x = -4, and then I need to divide both sides by 2 to get x = -2. So my two answers are -1 and -2. Now I am done.

Lesson Summary

Let's review. We learned that a quadratic trinomial is a quadratic expression with all three terms in the form of ax^2 + bx + c, where a, b, and c are numbers and not a 0. The method of factoring involves finding what multiplies together to get our quadratic. You will end up with two pairs of parentheses when you are done factoring.

The method involves writing down the factors of the first term and the last term. One pair of factors from the first term is then combined with a pair of factors from the second term so that when the combinations are multiplied and added together, they will get the middle term.

If the problem just says to factor, we can stop after filling in our two parentheses that result from factoring. If the problem says to solve, we then set each parentheses equal to 0 and then we solve for our variable. We will end up with two answers or solutions.

Summary:

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